Question

a thin 13 - cm - long solenoid has a total of 410 turns of wire and carries a current of 1.8 a. part a calculate the field inside the solenoid near the center. express your answer using two significant figures. b = t

Explanation:

Step1: Recall the formula for the magnetic - field inside a solenoid

The formula for the magnetic field inside a solenoid near the center is $B = \mu_0 nI$, where $\mu_0=4\pi\times 10^{- 7}\ T\cdot m/A$ is the permeability of free space, $n$ is the number of turns per unit length, and $I$ is the current.

Step2: Calculate the number of turns per unit length $n$

The length of the solenoid $L = 13\ cm=0.13\ m$ and the number of turns $N = 410$. The number of turns per unit length $n=\frac{N}{L}=\frac{410}{0.13}\ m^{-1}\approx3153.85\ m^{-1}$.

Step3: Calculate the magnetic field $B$

We know that $\mu_0 = 4\pi\times10^{-7}\ T\cdot m/A$ and $I = 1.8\ A$. Substitute $n$ and $I$ into the formula $B=\mu_0 nI$.
$B=(4\pi\times 10^{-7}\ T\cdot m/A)\times3153.85\ m^{-1}\times1.8\ A$.
$B = 4\pi\times10^{-7}\times3153.85\times1.8\ T$.
$B\approx7.1\times 10^{-3}\ T$.

Answer:

$7.1\times 10^{-3}\ T$